seratend
- 318
- 1
I’ve read this interesting discussion and I’d like to add the following comments.
Bell’s inequalities and more precisely EPR like states help in understanding how quantum states behave. There are many papers in arxiv about theses inequalities. Many of them show how classical statistics can locally break these inequalities even without the need to introduce local (statistical) errors in the experiment.
Here are 2 examples extracted form arxiv (far from being exhaustive).
Example 1 : quant-th/0209123 Laloë 2002 extensive paper on QM interpretation questions (to my opinion against local hidden variable theory, but open mind => lot of pointers and examples)
Example 2: quant-ph/0007005 Accardi 2000 (and later). An Example of how a classical probability space can break bell inequalities (contextual).
The approach of Nightlight, if I have correctly understood, is another way (I’ve missed it: thanks a lot for this new possibility): instead of breaking the inequalities, the “statistical errors” (some events not counted by the experiment, or the way the experiment data is calculated), if included in the final result, force the experiment to follow the bell inequalities. This is another point of view on what is “really” going on with the experiment.
All of these alternative examples use a classical probability space, i.e. the Kolmogorov axiomatization, where one take the adequate variables such that they can violate the Bell’s inequalities (and now, a way to enforce them).
Now, if the question is to know whether the bell’s inequalities experiments are relevant or not, one conservative approach is to try to know (at least feel, and the best, demonstrate), if in “general”, “sensible” experiments (quantum or classical or whatever we want) are most likely to break the bell’s inequalities or not. If the answer is no, then we must admit that aspect type of experiments have detected a rare event and that the leaving “statistical errors” seem not to help (in breaking the inequalities). If the answer is yes, well, we can say what we want :).
The papers against bell’s inequalities experiments, to my modest opinion, demonstrate that a sensible experiment is more likely to detect the inequalities breaking so that we can say what we want! That’s a little bit disappointing, because in this case we still not know if any quantum state may be described by any “local” classical probability space or not. I really prefer to get an good and solid explanation.
To end, I did not know before the Sica’s papers. But, I would like to understand the mechanism he (and Nightlight) used in order to force the Bell’s inequality matching. I follow Vanesh reasoning without problem, but the Nighlight one is a little bit more difficult to understand: where is the additional freedom used to enforce the inequality.
So, let's try to understand this problem in the special case of the well known Aspect et al experiment 1982, phys.rev. letters (where only very simple mathematics are used). I like to use a particular case before making a generalisation; it is easier to see where the problem is.
First let’s take 4 ideal discrete measurements (4 sets of data) of an Aspect type experiment with no lost sample during the measurement process.
If we take the classical expectations formulas with have :
S+= E(AB)+E(AB’)=1/N sum_i1 [A(i1)B(i1)]+ 1/N sum_i2 [A(i2)B’(i2)]
= 1/N sum_i1_i2[A(i1)B(i1)+ A(i2)B’(i2)] (1)
Where A(i1),B(i1) is the data collected by the first experiment and A(i2),B(i2) the data collected by the second experiment. With N --> ∞ (we also take the same sample number for each experiment).
In our particular case A(i1) is the result of the spin measurement of photon 1 on the A (same name as the observable) axis (+1 if spin |+>, -1 if spin |->) while B(i1) is the result of the spin measurement of photon 2 on the B axis (+1 if spin |+>, -1 if spin |->).
Each ideal measurement (given by label i1 or i2) thus gives two spin results (the two photons must be detected).
Etc … For the other measurement cases.
We thus have the second equation:
S-= E(A’B)-E(A’B’)=1/N sum_i3 [A’(i3)B(i3)]- 1/N sum_i4 [A’(i4)B’(i4)]
= 1/N sum_i3_i4[A(i3)B(i3)- A(i4)B(i4)] (2)
Labelling equation (1) or (2), ie, changing the ordering of label i1,i2,i3,i4 does not change the result (sum is commutative).
Now, If we want get the inequality S+=|E(AB)+E(AB’)|≤ 1+E(BB’), we first need to make a filter to the rhs equation (1), otherwise A cannot be factorized: we must select a subset of experiment samples with A(i1)=A(i2).
If we take a large samples number N, equation (1) is not changed with this filtering and we get:
|S+|= |E(AB)+E(AB’)|= 1/N |sum_i1_i2[A(i1)B(i1)+ A(i2)B’(i2)] |
= 1/N |sum_i1[A(i1)B(i1)+ A(i1)B’(i1)] |=
≤1/Nsum_i1 |[A(i1)B(i1)+ A(i1)B’(i1)]|
We then used the simple inequality |a.b+a.c|≤ 1+ a.c (|a|,|b|,|c| ≤1) for each label i1
|S+|= |E(AB)+E(AB’)| ≤1+1/N sum_i1[B(i1)B’(i1)] (3)
Remind that B’(i1) is the data of the second experiment relabelled with a subset of label i1. Now this re-labelling has a freedom because we may have several experiment results (50%) where A(i1)=A(i2).
So in equation (3) |sum_i1[B(i1)B’(i1)]]| depends on the artificial label order.
We also have almost the same inequality for equation (2)
|S+|= |E(A’B)-E(A’B’)|= 1/N |sum_i3_i4[A’(i3)B(i3)- A’(i4)B’(i4)] |
= 1/N |sum_i3[A’(i3)B(i3)- A’(i3)B’(i3)] |=
≤1/Nsum_i3 |A’(i3)B(i3)- A’(i3)B’(i3)|
We then used the simple inequality |a.b-a.c|≤ 1- a.c (|a|,|b|,|c| ≤1)
|S+|= |E(A’B)-E(A’B’)| ≤1-1/N sum_i3[B(i3)B’(i3)] (4)
So in equation (4) |sum_i1[B(i3)B’(i3)]]| depends on the artificial label ordering i3.
Now, we thus have the bell inequality:
|S=S++S-|≤ S++ S-= 2+1/N sum_i1_i3[B(i1)B’(i1)-B(i3)B’(i3)] (5)
where sum_i1_i3[B(i1)B’(i1)-B(i3)B’(i3)] depends on the labelling order we have used to filter and get this result.
I think that (3) and (4) may be the labelling order pb remarked by Nighlight in this special case.
Up to know, we have only spoken of collection of measurement results of values +1/-1.
Now if B is a random variable that depends only on the local experiment apparatus (the photon polarizer) we have B=B(apparatus_B, hv) where hv is the local hidden variable, we should have:
1/N.sum_i1[B(i1)B’(i1)= 1/N.sum_i3B(i3)B’(i3)] = <BB’> when N--> ∞.
(so we have the Bell inequality |S|≤2).
So, now I can use the Nighlight argument, ordering of B’(i1) and B’(i3) is totally artificial then the question is: should I got 1/N.sum_i1[B(i1)B’(i1)<> 1/N.sum_i3B(i3)B’(i3)] or the equality?
Moreover, Equation (5) seems to show that this kind of experiments are more likely to see a violation of a bell inequality as B(i1),B’(i2),B(i3)B’(i4) comes from 4 different experiments.
Seratend
P.S. Sorry if some minor mistakes are left.
Bell’s inequalities and more precisely EPR like states help in understanding how quantum states behave. There are many papers in arxiv about theses inequalities. Many of them show how classical statistics can locally break these inequalities even without the need to introduce local (statistical) errors in the experiment.
Here are 2 examples extracted form arxiv (far from being exhaustive).
Example 1 : quant-th/0209123 Laloë 2002 extensive paper on QM interpretation questions (to my opinion against local hidden variable theory, but open mind => lot of pointers and examples)
Example 2: quant-ph/0007005 Accardi 2000 (and later). An Example of how a classical probability space can break bell inequalities (contextual).
The approach of Nightlight, if I have correctly understood, is another way (I’ve missed it: thanks a lot for this new possibility): instead of breaking the inequalities, the “statistical errors” (some events not counted by the experiment, or the way the experiment data is calculated), if included in the final result, force the experiment to follow the bell inequalities. This is another point of view on what is “really” going on with the experiment.
All of these alternative examples use a classical probability space, i.e. the Kolmogorov axiomatization, where one take the adequate variables such that they can violate the Bell’s inequalities (and now, a way to enforce them).
Now, if the question is to know whether the bell’s inequalities experiments are relevant or not, one conservative approach is to try to know (at least feel, and the best, demonstrate), if in “general”, “sensible” experiments (quantum or classical or whatever we want) are most likely to break the bell’s inequalities or not. If the answer is no, then we must admit that aspect type of experiments have detected a rare event and that the leaving “statistical errors” seem not to help (in breaking the inequalities). If the answer is yes, well, we can say what we want :).
The papers against bell’s inequalities experiments, to my modest opinion, demonstrate that a sensible experiment is more likely to detect the inequalities breaking so that we can say what we want! That’s a little bit disappointing, because in this case we still not know if any quantum state may be described by any “local” classical probability space or not. I really prefer to get an good and solid explanation.
To end, I did not know before the Sica’s papers. But, I would like to understand the mechanism he (and Nightlight) used in order to force the Bell’s inequality matching. I follow Vanesh reasoning without problem, but the Nighlight one is a little bit more difficult to understand: where is the additional freedom used to enforce the inequality.
So, let's try to understand this problem in the special case of the well known Aspect et al experiment 1982, phys.rev. letters (where only very simple mathematics are used). I like to use a particular case before making a generalisation; it is easier to see where the problem is.
First let’s take 4 ideal discrete measurements (4 sets of data) of an Aspect type experiment with no lost sample during the measurement process.
If we take the classical expectations formulas with have :
S+= E(AB)+E(AB’)=1/N sum_i1 [A(i1)B(i1)]+ 1/N sum_i2 [A(i2)B’(i2)]
= 1/N sum_i1_i2[A(i1)B(i1)+ A(i2)B’(i2)] (1)
Where A(i1),B(i1) is the data collected by the first experiment and A(i2),B(i2) the data collected by the second experiment. With N --> ∞ (we also take the same sample number for each experiment).
In our particular case A(i1) is the result of the spin measurement of photon 1 on the A (same name as the observable) axis (+1 if spin |+>, -1 if spin |->) while B(i1) is the result of the spin measurement of photon 2 on the B axis (+1 if spin |+>, -1 if spin |->).
Each ideal measurement (given by label i1 or i2) thus gives two spin results (the two photons must be detected).
Etc … For the other measurement cases.
We thus have the second equation:
S-= E(A’B)-E(A’B’)=1/N sum_i3 [A’(i3)B(i3)]- 1/N sum_i4 [A’(i4)B’(i4)]
= 1/N sum_i3_i4[A(i3)B(i3)- A(i4)B(i4)] (2)
Labelling equation (1) or (2), ie, changing the ordering of label i1,i2,i3,i4 does not change the result (sum is commutative).
Now, If we want get the inequality S+=|E(AB)+E(AB’)|≤ 1+E(BB’), we first need to make a filter to the rhs equation (1), otherwise A cannot be factorized: we must select a subset of experiment samples with A(i1)=A(i2).
If we take a large samples number N, equation (1) is not changed with this filtering and we get:
|S+|= |E(AB)+E(AB’)|= 1/N |sum_i1_i2[A(i1)B(i1)+ A(i2)B’(i2)] |
= 1/N |sum_i1[A(i1)B(i1)+ A(i1)B’(i1)] |=
≤1/Nsum_i1 |[A(i1)B(i1)+ A(i1)B’(i1)]|
We then used the simple inequality |a.b+a.c|≤ 1+ a.c (|a|,|b|,|c| ≤1) for each label i1
|S+|= |E(AB)+E(AB’)| ≤1+1/N sum_i1[B(i1)B’(i1)] (3)
Remind that B’(i1) is the data of the second experiment relabelled with a subset of label i1. Now this re-labelling has a freedom because we may have several experiment results (50%) where A(i1)=A(i2).
So in equation (3) |sum_i1[B(i1)B’(i1)]]| depends on the artificial label order.
We also have almost the same inequality for equation (2)
|S+|= |E(A’B)-E(A’B’)|= 1/N |sum_i3_i4[A’(i3)B(i3)- A’(i4)B’(i4)] |
= 1/N |sum_i3[A’(i3)B(i3)- A’(i3)B’(i3)] |=
≤1/Nsum_i3 |A’(i3)B(i3)- A’(i3)B’(i3)|
We then used the simple inequality |a.b-a.c|≤ 1- a.c (|a|,|b|,|c| ≤1)
|S+|= |E(A’B)-E(A’B’)| ≤1-1/N sum_i3[B(i3)B’(i3)] (4)
So in equation (4) |sum_i1[B(i3)B’(i3)]]| depends on the artificial label ordering i3.
Now, we thus have the bell inequality:
|S=S++S-|≤ S++ S-= 2+1/N sum_i1_i3[B(i1)B’(i1)-B(i3)B’(i3)] (5)
where sum_i1_i3[B(i1)B’(i1)-B(i3)B’(i3)] depends on the labelling order we have used to filter and get this result.
I think that (3) and (4) may be the labelling order pb remarked by Nighlight in this special case.
Up to know, we have only spoken of collection of measurement results of values +1/-1.
Now if B is a random variable that depends only on the local experiment apparatus (the photon polarizer) we have B=B(apparatus_B, hv) where hv is the local hidden variable, we should have:
1/N.sum_i1[B(i1)B’(i1)= 1/N.sum_i3B(i3)B’(i3)] = <BB’> when N--> ∞.
(so we have the Bell inequality |S|≤2).
So, now I can use the Nighlight argument, ordering of B’(i1) and B’(i3) is totally artificial then the question is: should I got 1/N.sum_i1[B(i1)B’(i1)<> 1/N.sum_i3B(i3)B’(i3)] or the equality?
Moreover, Equation (5) seems to show that this kind of experiments are more likely to see a violation of a bell inequality as B(i1),B’(i2),B(i3)B’(i4) comes from 4 different experiments.
Seratend
P.S. Sorry if some minor mistakes are left.